21

有没有一种快速的方法可以将纬度和经度坐标转换为 R 中的州代码？我一直在使用 zipcode 包作为查找表，但是当我查询很多 lat/long 值时它太慢了

如果不在 R 中，有没有办法使用谷歌地理编码器或任何其他类型的快速查询服务来做到这一点？

谢谢！

4

5 回答 5

45

这里有两种选择，一种使用sf，一种使用sp包函数。sf是用于分析空间数据的更现代的（并且在 2020 年推荐）包，但如果它仍然有用，我将留下我在 2012 年的原始答案，展示如何使用与sp相关的函数来做到这一点。

方法1（使用sf）：

library(sf)
library(spData)

## pointsDF: A data.frame whose first column contains longitudes and
##           whose second column contains latitudes.
##
## states:   An sf MULTIPOLYGON object with 50 states plus DC.
##
## name_col: Name of a column in `states` that supplies the states'
##           names.
lonlat_to_state <- function(pointsDF,
                            states = spData::us_states,
                            name_col = "NAME") {
    ## Convert points data.frame to an sf POINTS object
    pts <- st_as_sf(pointsDF, coords = 1:2, crs = 4326)

    ## Transform spatial data to some planar coordinate system
    ## (e.g. Web Mercator) as required for geometric operations
    states <- st_transform(states, crs = 3857)
    pts <- st_transform(pts, crs = 3857)

    ## Find names of state (if any) intersected by each point
    state_names <- states[[name_col]]
    ii <- as.integer(st_intersects(pts, states))
    state_names[ii]
}

## Test the function with points in Wisconsin, Oregon, and France
testPoints <- data.frame(x = c(-90, -120, 0), y = c(44, 44, 44))
lonlat_to_state(testPoints)
## [1] "Wisconsin" "Oregon"    NA

如果您需要更高分辨率的状态边界，请使用或通过其他方式将您自己的矢量数据作为sf对象读取。sf::st_read()一个不错的选择是安装rnaturalearth包并使用它从rnaturalearthhires加载状态向量层。然后使用lonlat_to_state()我们刚刚定义的函数，如下所示：

library(rnaturalearth)
us_states_ne <- ne_states(country = "United States of America",
                          returnclass = "sf")
lonlat_to_state(testPoints, states = us_states_ne, name_col = "name")
## [1] "Wisconsin" "Oregon"    NA

要获得非常准确的结果，您可以从此页面下载包含GADM 维护的美国行政边界的地理包。然后，加载状态边界数据并像这样使用它们：

USA_gadm <- st_read(dsn = "gadm36_USA.gpkg", layer = "gadm36_USA_1")
lonlat_to_state(testPoints, states = USA_gadm, name_col = "NAME_1")
## [1] "Wisconsin" "Oregon"    NA

方法2（使用sp）：

这是一个函数，它在较低的 48 个状态中获取 lat-longs 的 data.frame，并且对于每个点，返回它所在的状态。

大多数函数只是在包中准备函数所需的SpatialPoints和SpatialPolygons对象，它完成了计算点和多边形的“交点”的真正繁重的工作：over()sp

library(sp)
library(maps)
library(maptools)

# The single argument to this function, pointsDF, is a data.frame in which:
#   - column 1 contains the longitude in degrees (negative in the US)
#   - column 2 contains the latitude in degrees

lonlat_to_state_sp <- function(pointsDF) {
    # Prepare SpatialPolygons object with one SpatialPolygon
    # per state (plus DC, minus HI & AK)
    states <- map('state', fill=TRUE, col="transparent", plot=FALSE)
    IDs <- sapply(strsplit(states$names, ":"), function(x) x[1])
    states_sp <- map2SpatialPolygons(states, IDs=IDs,
                     proj4string=CRS("+proj=longlat +datum=WGS84"))

    # Convert pointsDF to a SpatialPoints object 
    pointsSP <- SpatialPoints(pointsDF, 
                    proj4string=CRS("+proj=longlat +datum=WGS84"))

    # Use 'over' to get _indices_ of the Polygons object containing each point 
        indices <- over(pointsSP, states_sp)

    # Return the state names of the Polygons object containing each point
    stateNames <- sapply(states_sp@polygons, function(x) x@ID)
    stateNames[indices]
}

# Test the function using points in Wisconsin and Oregon.
testPoints <- data.frame(x = c(-90, -120), y = c(44, 44))

lonlat_to_state_sp(testPoints)
[1] "wisconsin" "oregon" # IT WORKS
于 2012-01-06T00:39:24.740 回答
12

你可以用几行 R 来完成。

library(sp)
library(rgdal)
#lat and long
Lat <- 57.25
Lon <- -9.41
#make a data frame
coords <- as.data.frame(cbind(Lon,Lat))
#and into Spatial
points <- SpatialPoints(coords)
#SpatialPolygonDataFrame - I'm using a shapefile of UK counties
counties <- readOGR(".", "uk_counties")
#assume same proj as shapefile!
proj4string(points) <- proj4string(counties)
#get county polygon point is in
result <- as.character(over(points, counties)$County_Name)
于 2013-07-27T10:10:11.440 回答
3

参见?oversp 包。

您需要将状态边界设置为SpatialPolygonsDataFrame。

于 2012-01-06T00:02:28.933 回答
0

示例数据（多边形和点）

library(raster)
pols <- shapefile(system.file("external/lux.shp", package="raster"))
xy <- coordinates(p)

使用 raster::extract

extract(p, xy)

#   point.ID poly.ID ID_1       NAME_1 ID_2           NAME_2 AREA
#1         1       1    1     Diekirch    1         Clervaux  312
#2         2       2    1     Diekirch    2         Diekirch  218
#3         3       3    1     Diekirch    3          Redange  259
#4         4       4    1     Diekirch    4          Vianden   76
#5         5       5    1     Diekirch    5            Wiltz  263
#6         6       6    2 Grevenmacher    6       Echternach  188
#7         7       7    2 Grevenmacher    7           Remich  129
#8         8       8    2 Grevenmacher   12     Grevenmacher  210
#9         9       9    3   Luxembourg    8         Capellen  185
#10       10      10    3   Luxembourg    9 Esch-sur-Alzette  251
#11       11      11    3   Luxembourg   10       Luxembourg  237
#12       12      12    3   Luxembourg   11           Mersch  233
于 2018-10-17T01:08:48.580 回答
-1

使用非常简单sf：

library(maps)
library(sf)

## Get the states map, turn into sf object
US <- st_as_sf(map("state", plot = FALSE, fill = TRUE))

## Test the function using points in Wisconsin and Oregon
testPoints <- data.frame(x = c(-90, -120), y = c(44, 44))

# Make it a spatial dataframe, using the same coordinate system as the US spatial dataframe
testPoints <- st_as_sf(testPoints, coords = c("x", "y"), crs = st_crs(US))

#.. and perform a spatial join!
st_join(testPoints, US)


         ID        geometry
1 wisconsin  POINT (-90 44)
2    oregon POINT (-120 44)
于 2018-10-17T01:39:23.067 回答